Question

Evaluate: $\int x\frac{d}{dx}\left(\frac{x^2+1}{x}\right)dx$

• A. $\frac{x^3}{2}-\ln x+c$
• B. $\frac{x^2}{2}-\ln x+c$
• C. $\frac{x^2}{2}-\ln (x^2+1)+c$
• D. $\frac{(x+1{)}^2}{2}-\ln x+c$

Answer 1

The correct option is B $\frac{x^2}{2}-\ln x+c$

$u\frac{dv}{dx}=\frac{duv}{dx}-\frac{vdu}{dx}u=xv=\frac{x^2+1}{x}x\frac{d}{dx}\left(\frac{x^2+1}{x}\right)=\frac{d}{dx}\left(x\frac{x^2+1}{x}\right)-\frac{x^2+1}{x}\frac{dx}{dx}I=\int x\frac{d}{dx}\left(\frac{x^2+1}{x}\right)dx=\int [\frac{d}{dx}\left(x\frac{x^2+1}{x}\right)-\frac{x^2+1}{x}\frac{dx}{dx}]dxI=\int \frac{d}{dx}(x^2+1)dx-\int \frac{x^2+1}{x}dxI=(x^2+1)-\int (x+\frac{1}{x})dx+KI=x^2+1-[\frac{x^2}{2}+\ln x]+C^{\prime }I=x^2-\frac{x^2}{2}-\ln x\{C=C^{\prime }+1\}I=\frac{x^2}{2}-\ln x+C$